mk lm mẫu cho bạn 1 phần nhé

a) \(A=3x^2+y^2+10x-2xy+26\)

\(=\left(x^2-2xy+y^2\right)+2\left(x^2+5x+6,25\right)+13,5\)

\(=\left(x-y\right)^2+2\left(x+2,5\right)^2+13,5\ge13,5\)

Dấu "=" xảy ra <=>  \(x=y=-2,5\)

Vậy MIN A = 13,5  khi  x = y = - 2,5

Cảm ơn Đường Quỳnh Giang nhiều nhé😊

Đường Quỳnh Giang ơi, làm cho mình các phần C, D, E nhé 😊

a, +/  Có  \(A=4x-x^2+3=4x-x^2+4-1\)

                     \(=-\left(-2.2x+x^2+2^2\right)+1=1-\left(x-2\right)^2\)

            do  \(\left(x-2\right)^2\ge0\forall x\in R\Rightarrow A\le1\)

                          \(\Rightarrow maxA=1\)tại  \(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

                            Vậy max A=1 tại x=2

      +/ Có \(B=x-x^2=2.\frac{1}{2}x-x^2-\frac{1}{4}+\frac{1}{4}\)

                     \(=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{4}=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\)

              \(\Rightarrow A\le\frac{1}{4}\)do\(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow maxB=\frac{1}{4}\)tại \(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

              Vậy max B =\(\frac{1}{4}\)tại  x=\(\frac{1}{2}\)

Lời giải:
Đặt biểu thức trên là $A$

$A=4x^2+4x-12xy-2y+10y^2+8$

$=(4x^2-12xy+9y^2)+4x-2y+y^2+8$

$=(2x-3y)^2+2(2x-3y)+4y+y^2+8$

$=(2x-3y)^2+2(2x-3y)+1+(y^2+4y+4)+3$

$=(2x-3y+1)^2+(y+2)^2+3\geq 0+0+3=3$

Vậy $A_{\min}=3$. Giá trị này đạt tại $2x-3y+1=y+2=0$

$\Leftrightarrow y=-2; x=\frac{-7}{2}$

\(A=3\left(4x^2-4xy+y^2\right)-10\left(2x-y\right)+8\)

\(A=3\left(2x-y\right)^2-10\left(2x-y\right)+8\)

\(A=\left(6x-3y-4\right)\left(2x-y-2\right)\)

a) Ta có: \(\left(2x+7\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\left(2x+7\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(2x+7-x-3\right)\left(2x+7+x+3\right)=0\)

\(\Leftrightarrow\left(x+4\right)\cdot\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-4;-\dfrac{10}{3}\right\}\)

b) Ta có: \(\left(4x+14\right)^2=\left(7x+2\right)^2\)

\(\Leftrightarrow\left(4x+14\right)^2-\left(7x+2\right)^2=0\)

\(\Leftrightarrow\left(4x+14-7x-2\right)\left(4x+14+7x+2\right)=0\)

\(\Leftrightarrow\left(-3x+12\right)\left(11x+16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x+12=0\\11x+16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-12\\11x=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{16}{11}\end{matrix}\right.\)Vậy: \(S=\left\{4;-\dfrac{16}{11}\right\}\)

(2x+7)²=(x+3)²

=>(2x+7)²-(x+3)²=0

=>(2x+7-x-3)(2x+7+x+3)=0

=>(x-4)(3x+10)=0

=>x-4=0 hoặc 3x+10=0

TH1:x-4=0=>x=4

TH2:3x+10=0=>x=-10/3

(4x+14)²=(7x+2)²

(4x+14)²-(7x+2)²=0

(4x+14-7x-2)(4x+14+7x+2)=0

(-3x+12)(11x+16)=0

TH1:-3x+12=0=>x=4

TH2:11x+16=0=>x=-16/11

Đề bài là gì bạn nhỉ ?